we can find the density of Y by integrating the joint density with respect to x

thus getting

fy(Y) = (1/6)*(y^3)*e^(-y)

now we are interested in finding the pdf of z = g(y) = y^3, g(y) is a monotonic function of y thus the resulting density function is given by:

fz(Z) = abs(1 / g'(g-1(z)) )*fy(g-1(z))

where g-1 is the inverse of g in our case g-1(z) = z^(1/3), thus we get:

fz(Z) = [(1/3)*Z^(-2/3)]*(1/6)*Z*e^(-Z^(1/3)) =

[(1/18)*Z^(1/3)]*e^(-Z^(1/3)) for 0 <= Z <= inf